Conway Notation specification does not specify that beveling can only be done by vertex valances. I think this may have been an oversight! Beveling is actually truncation by 1/2. Therefore since Conway Notation allows for truncation by vertex valance, so should it be with Beveling. I am going to pr...

But I'm sure there's a more straight forward way. Actually, it's a rectified truncated octahedron I suppose. It can be found by truncating the on 3-valance vertices of the truncated octahedron by a factor of 1/2. It can also be found by beveling only on the 3-valance vertices in conway notation whi...

There are many that can't work. For example if you augment a cube with cubes it will be non-convex and is already equilateral.

That models seems familiar. How did you create it?

I haven't been able to make one with the rhombic enneacontahedron truncated on the 5-valance vertices. It works on the 3 and 6 valance vertices, but on the 5 it collapses into non-convex.

Roger

For the step 2's there seems to be a kind of order. Such that it might be possible to find the pattern to make the cells. Then there could be a way to generate the Nth model.

For the other steps it would be another matter.

Roger

Very intersting. I made one observation. Of all of the step tegums you have sent (thanks!) I noticed that for each N, at least one of them has a similar cell type. For an N-choron is taking a wedge of an 3D (N-2).4.4 prism. e.g. for the Heptachoron the cell is the wedge of a 5.4.4 prism. The Octacho...

How do you do this?! It looks like there is no other way to make these than by hand. It is amazing to me that such an oddball looking outer cell can contain copies of itself inside. Do you have the stel file for this? I had thought I might try to construct what you had just by looking but I'd have t...

Jabe wrote:the 8 - 2 looks totally different.

Not trying to be a lamer here, and I really tried to see if this was out there some where. In any search I would ultimately land on your 4D dice page!

I'm convinced you may be the only person in the world who has seen it.

Roger

Since it is systematic I wonder if these could be generated. If the connectivity could be systemized it could. You mentioned that there are two 8-cells. Try to look something like that up on google! If there exists information on an 8-cell it is inevitably the tesseract. Looking at what you've writt...

Jabe wrote:These exist for all N greater than 4, for N=5 we get the pentachoron.
For N=6 we get the triangle duoprism, for N=8 there are two and one is the tesseract.

And the number of configurations increases as N grows larger I would presume. Is there a formula for that?

Roger

I don't see this as a problem but some of the Johnson solids allow "Create Unform Polychoron from 3D vertex figure". Not a lot of them but a few. They result in something that is not closed, but creates some very interesting results anyway. J1 - no complaints but says it has 8 cells but I only see 7...

I'm aware of two potential hendecachora with identical cells that could be modeled in 4D, however they aren't the one with the hemi-icosahedra - that one requires some strange curved 4-space. I have yet to model the two 11-celled polychora, but I have done the 7-cell, two 9-cells, and three distinc...

I don't really understand the shape you're after. Carlos' paper is on line. It shows ways of trying to display the 11-cell but using purely 3D constructs. i.e usings 11 vertices and so on. I was bring such models into Stella but I had to double the number of faces (replicated on top of one another)...

Maybe the fact that those coplanar faces also pass through the centre of the model makes things worse. Rob. Carlo's models always were tweaked so that the intersection didn't happen. I can try that. I remember some of profesional geometers saying that the 11-cell and 57-cell can't be represented in...

Hi, The 11-cell or hendecachoron is 11 hemi-icosahedra. I have a hemi-icosahedra (OFF file below). I was wondering if the 11-cell might be attainable (in Stella4D, pressing "u") but it doesn't appear to be. It stops at 10 cells and can't display anything. Is this because the hemi-icosahedra is non-o...